I'm going through a practice exam which has solutions and I'm a bit confused about this question:
Write down the third order Taylor series method for the differential equation
$y'(x)=-y+x+1, \ \ 0\le x\le1, \ \ y(0)=1.$
Compute three steps using $h=0.1$.
The solutions computes derivatives as $ \displaystyle f(x,y)=-y+x+1, \frac{d}{dx}f(x,y)=y-x,\frac{d^2}{dx^2}f(x,y)=-y+x.$
I don't understand how on earth this was achieved. Is there a different type of differentiation at play here?
When I did it I obtained $ \displaystyle f(x,y)=-y+x+1, \frac{d}{dx}f(x,y)=-y'+1,\frac{d^2}{dx^2}f(x,y)=-y''$. What am I missing here?
You know that $$y'=-y+x+1$$ So you found that $$\frac{d}{dx}f(x,y)=-y'+1$$ You can substitute, which will give you $$y'=y-x$$ Same with the second derivative $(f''=y')$.